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%TCIDATA{Created=Wed Jul 28 10:29:31 1999}
%TCIDATA{LastRevised=Thu May 25 15:03:33 2000}
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\begin{document}
\title[Hyponormality of Toeplitz Operators]{Kernels of Hankel Operators and Hyponormality of Toeplitz Operators.}
\author{Caixing Gu}
\address{California Polytechnic State University\\
San Luis Obispo}
\email{cgu@calpoly.edu}
\author{Jonathan E. Shapiro}
\email{jshapiro@calpoly.edu}
\urladdr{http://www.calpoly.edu/\symbol{126}jshapiro}
\date{September 17, 1999}
\subjclass{Primary 47-B35, 47-B20}
\keywords{Hankel Operator, Toeplitz Operator, Hyponormality}
\begin{abstract}
We give a formula for $\ker H_{\overline{\theta_{1}}}^{\ast}H_{\overline
{\theta_{2}}}$ and describe when $\ker H_{\overline{\theta_{1}}}^{\ast
}H_{\overline{\theta_{2}}}=\ker H_{\overline{\theta_{2}}}$. We explore the
hyponormality of Toeplitz operators whose symbols are of circulant type and
some more general types. In addition, we discuss formulas for and estimates of
the rank of the self-commutator of a hyponormal Toeplitz operator.
\end{abstract}\maketitle
\section{Introduction}
Brown and Halmos \cite{BH} started the study of the algebraic properties of
Toeplitz operators. They showed that the Toeplitz operator $T_{\varphi}$ with
symbol $\varphi$ is normal if and only if $\varphi=\alpha u+\beta$ for some
real-valued function $u$ and complex numbers $\alpha$ and $\beta$. Cowen
\cite{cowenh} recently showed that $T_{\varphi}$ is hyponormal if and only if
there exists $k\in H^{\infty}$,\emph{\ }$\left\| k\right\| _{\infty}\leq
1$,$\ $such that $\ \varphi_{-}-k\overline{\varphi_{+}}\in H^{2}$, where
$\varphi_{+}=P(\varphi)$ and $\varphi_{-}=(I-P)(\varphi).$ See also Nakazi and
K. Takahashi \cite{nt} and the first named author \cite{gug} for
generalizations and refinements of this characterization.
Recent efforts have been made to give more explicit conditions for the
hyponormality of $T_{\varphi}.$ That is, how we can actually check if there is
such a $k\in H^{\infty}$,\emph{\ }$\left\| k\right\| _{\infty}\leq
1$\emph{\ \ }$\ $such that $\ \varphi_{-}-k\overline{\varphi_{+}}\in H^{2}.$
\ Particular attention has been paid to Toeplitz operators with polynomial
symbols. In Particular, using the results of Nakazi and Takahashi \cite{nt},
and Zhu \cite{zhu}, Lee and his collaborators \cite{flt}, \cite{FL},
\cite{lee1} and \cite{lee2} gave explicit conditions for the hyponormality of
Toeplitz operators with polynomial symbols whose coefficients satisfy certain
symmetric or circulant relations. The main purpose of this paper is to put
their results in a much more general and abstract setting and to give explicit
criteria for the hyponormality of\ a broader class of Toeplitz operators. Our
approach emphasizes the use of Hankel operators and the proofs are done in a
more abstract and often simpler way.
To be more specific, we now outline the plan of the paper. Some questions
about the hyponormality of Toeplitz operators are seen to be related to
properties of the kernels of Hankel operators and products of Hankel
operators. In Section \ref{hankelkernelsec}, we explore the kernels of
products of Hankel operators, leading to the explicit formula for $\ker
H_{\overline{\theta_{1}}}^{\ast}H_{\overline{\theta_{2}}}$. We use this
formula to understand better when $\ker H_{\overline{\theta_{1}}}^{\ast
}H_{\overline{\theta_{2}}}=\ker H_{\overline{\theta_{2}}}$ and to provide a
simple proof of the main result of Nakazi in \cite{N}.
In Section \ref{hypofsec}, we use the result mentioned above to tell us about
the hyponormality of a Toeplitz operator with circulant-type symbol. \ This
allows us very easily to derive the explicit conditions for hyponormality of a
Toeplitz operator when the symbol is a circulant polynomial.
The results of Section \ref{hypofsec} concerning hyponormality of Toeplitz
operators with circulant-type symbols are generalized to a broader class of
symbols in Section \ref{hypopsec}. Formulas for and estimates of the rank of
the self-commutator of a hyponormal Toeplitz operator are then developed in
Section \ref{ranksec}.
For Toeplitz operators with symbols of bounded type, we reduce the
determination of hyponormality to the computation of the norm of certain
Hankel operators. This takes place in Section \ref{moresec}. \
Finally, in Section \ref{normsec}, we use the computation of Hankel operator
norms to give more explicit examples of hyponormal Toeplitz operators with
symbols satisfying some symmetric conditions. It is a remarkable fact that the
Hankel operators we are interested in here have been studied extensively in
the recent literature because of their important application in robust control
theory; see the book \cite{foias}\ by Foias, \"{O}zbay and Tannenbaum for
details and more references. In particular, we indicate an efficient algorithm
from \cite{ozbay} by the first named author, Toker and \"{O}zbay for the
computation of Hankel operator norms. We illustrate this algorithm by a
self-contained discussion in a simple case. We then show how some conditions
on the coefficients of the polynomial symbols of hyponormal Toeplitz operators
in \cite{flt} and \cite{lee2} follow from this simple case, and we conclude
the paper with an explicit example of hyponormal Toeplitz operators with
irrational symbols.
\subsection{Notation and Preliminary Results}
For $\varphi$ in $L^{\infty}$, the Toeplitz operator $T_{\varphi}%
:H^{2}\rightarrow H^{2}$ is defined by $T_{\varphi}h=P\left( \varphi
h\right) $, where $P$ is the projection operator from $L^{2}$ to $H^{2}$. Let
$J:\left( H^{2}\right) ^{\perp}\rightarrow H^{2}$ be given by $Je^{-in\theta
}=e^{i(n-1)\theta}$ for $n\geq1$. The Hankel operator $H_{f}:H^{2}\rightarrow
H^{2}$ will be defined by $H_{f}h=J\left( I-P\right) \left( fh\right) $.
Let $S=T_{z}$ be the shift. It is then true that $H_{f}S=S^{\ast}H_{f}$. Also,
for $f\in L^{\infty}$, we define $\widetilde{f}=\overline{f\left(
\overline{z}\right) }$, and note that $H_{f}^{\ast}=H_{\widetilde{f}}$. \ We
can then verify, using the definition of the Hankel operator, that, for $g\in
H^{\infty}$,
\begin{equation}
H_{f}T_{g}=H_{fg}=T_{\widetilde{g}}^{\ast}H_{f}\text{.}\label{hid}%
\end{equation}
Using (\ref{hid}) and the connection between Hankel and Toeplitz operators
\[
T_{\varphi\psi}-T_{\varphi}T_{\psi}=H_{\overline{\varphi}}^{\ast}H_{\psi
}\text{, \ where }\varphi,\psi\in L^{\infty}\text{,}%
\]
we can show
\begin{lemma}
\label{fromgu1}For an inner function $\theta$ and $\varphi\in L^{\infty}$,
\[
H_{\varphi}^{\ast}H_{\varphi}-H_{\theta\varphi}^{\ast}H_{\theta\varphi
}=H_{\varphi}^{\ast}\left( 1-T_{\widetilde{\theta}}T_{\widetilde{\theta}%
}^{\ast}\right) H_{\varphi}=H_{\varphi}^{\ast}H_{\overline{\widetilde{\theta
}}}^{\ast}H_{\overline{\widetilde{\theta}}}H_{\varphi}=H_{\varphi}^{\ast
}H_{\overline{\theta}}H_{\overline{\theta}}^{\ast}H_{\varphi}.
\]
\end{lemma}
For an inner function $\theta$, $\mathcal{H}(\theta)=H^{2}\ominus\theta H^{2}%
$. \ We will make use of the characterization of $\mathcal{H}(\theta)$:
\[
\mathcal{H}(\theta)=\left\{ f\in H^{2}|\overline{\theta}f\in\overline
{H_{0}^{2}}\right\} \text{.}%
\]
It is then easy to show
\begin{lemma}
\label{hthetalemma}For an inner function $\theta$, $\overline{z}%
\theta\overline{\mathcal{H}(\theta)}=\mathcal{H}(\theta).$
\end{lemma}
\begin{proof}
If $f\in\mathcal{H}(\theta)$, then $\overline{z}\theta\overline{f}%
\in\mathcal{H}(\theta)$ since $\overline{\theta}\overline{z}\theta\overline
{f}=\overline{z}\overline{f}\in\overline{H_{0}^{2}}$, thus $\overline{z}%
\theta\overline{\mathcal{H}(\theta)}\subset\mathcal{H}(\theta)$. \ From this
inclusion, we can multiply both sides by $z\overline{\theta}$ and take
conjugates to get $\mathcal{H}(\theta)\subset\overline{z}\theta\overline
{\mathcal{H}(\theta)}$.
\end{proof}
We will also frequently make use of the facts that $\ker H_{\overline{\theta}%
}=\theta H^{2}$, and thus $\operatorname*{range}\left( H_{\overline{\theta}%
}^{\ast}\right) =\mathcal{H}(\theta)$.
\section{Hankel Operator Kernels\label{hankelkernelsec}}
It has been shown by the first named author that
\begin{theorem}
[Gu, 1999]\label{kernelthm}For any two $L^{\infty}$ functions $f$ and $g$,
\[
\text{either \ }\ker H_{f}^{\ast}H_{g}=\ker H_{g}\,\,\text{or \ }\ker
H_{g}^{\ast}H_{f}=\ker H_{f}\text{.}%
\]
\end{theorem}
See \cite{GU2} for a proof of this theorem and more general results for
products of finite many Hankel operators. As a corollary of this theorem, we
get a result of Axler, Chang, and Sarason in \cite{ACS}, which can be stated as
\begin{theorem}
[Axler, Chang, and Sarason, 1978]$H_{f}^{\ast}H_{g}$ has finite rank exactly
when at least one of $\left( I-P\right) f$ and $\left( I-P\right) g$ is a
rational function. In this case, the rank of $H_{f}^{\ast}H_{g}$ is the
minimum of the degrees of $\left( I-P\right) f$ and $\left( I-P\right) g$.
\end{theorem}
\begin{proof}
$H_{f}^{\ast}H_{g}$ has finite rank iff $H_{g}^{\ast}H_{f}$ does, too, and by
Theorem \ref{kernelthm}, this rank, which is the codimension of the kernel, is
either the rank of $H_{g}$ or $H_{f}$, at least one of which is finite. Since
the rank of $H_{g}^{\ast}H_{f}$ is at most the rank of $H_{f}$ and the rank of
$H_{f}^{\ast}H_{g}$ (which must be the same) is at most the rank of $H_{g}$,
this rank must be the minimum of the ranks of $H_{g}$ and $H_{f}$, which is
the minimum of the degrees of $\left( I-P\right) f$ and $\left( I-P\right)
g$.
\end{proof}
We now wish to understand better the when $\ker H_{f}^{\ast}H_{g}=\ker H_{g} $
and when $\ker H_{g}^{\ast}H_{f}=\ker H_{f}$, particularly in the case where
$f$ and $g$ are conjugates of inner functions.
\begin{theorem}
\label{hkp2}For any two inner functions $\theta_{1}$ and $\theta_{2}$,
\[
\ker H_{\overline{\theta_{1}}}^{\ast}H_{\overline{\theta_{2}}}=\theta_{2}%
H^{2}\oplus\theta_{2}\overline{z}\overline{\left( \theta_{1}H^{2}%
\cap\mathcal{H}\left( \theta_{2}\right) \right) }\text{.}%
\]
\end{theorem}
\begin{proof}
$\ker H_{\overline{\theta_{1}}}^{\ast}H_{\overline{\theta_{2}}}\supset\ker
H_{\overline{\theta_{2}}}=\theta_{2}H^{2}$, so for any $p$ in $\ker
H_{\overline{\theta_{1}}}^{\ast}H_{\overline{\theta_{2}}}$ which is not in
$\theta_{2}H^{2}$, $p\in\ker H_{\overline{\theta_{1}}}^{\ast}H_{\overline
{\theta_{2}}}$ is equivalent to $0\neq H_{\overline{\theta_{2}}}p\in\ker
H_{\overline{\theta_{1}}}^{\ast}=J\left( \overline{z}\overline{\theta_{1}%
}\overline{H^{2}}\right) ,$ or
\[
\overline{\theta_{2}}p=\overline{z}\overline{\theta_{1}}\overline{h}+k
\]
for some $H^{2}$ functions $h$ and $k$, with $h\neq0$. We rewrite this as
\[
p=\overline{z}\overline{\theta_{1}}\overline{h}\theta_{2}+\theta_{2}k\text{.}%
\]
Now consider the orthogonal projection of $p$ onto the space $\mathcal{H}%
\left( \theta_{2}\right) $. This projection is given by $\overline
{z}\overline{\theta_{1}}\overline{h}\theta_{2}$, since $\overline{z}%
\overline{\theta_{1}}\overline{h}\theta_{2}$ can be seen to be in
$\mathcal{H}\left( \theta_{2}\right) $, while $\theta_{2}k$ is orthogonal to
$\mathcal{H}\left( \theta_{2}\right) $. This tells us that $\ker
H_{\overline{\theta_{1}}}^{\ast}H_{\overline{\theta_{2}}}$ consists of the
orthogonal sum of $\theta_{2}H^{2}$ and elements of the form $\overline
{z}\overline{\theta_{1}}\overline{h}\theta_{2}$ which are in $H^{2}$, i.e.,
\[
\ker H_{\overline{\theta_{1}}}^{\ast}H_{\overline{\theta_{2}}}=\theta_{2}%
H^{2}\oplus\theta_{2}\overline{z}\overline{\theta_{1}}\overline{H^{2}}\cap
H^{2}\text{.}%
\]
Since $\theta_{2}\overline{z}\overline{\theta_{1}}\overline{H^{2}}%
\subset\mathcal{H}\left( \theta_{2}\right) $, we can write
\begin{align*}
\ker H_{\overline{\theta_{1}}}^{\ast}H_{\overline{\theta_{2}}} & =\theta
_{2}H^{2}\oplus\theta_{2}\overline{z}\overline{\theta_{1}}\overline{H^{2}}%
\cap\mathcal{H}\left( \theta_{2}\right) \\
& =\theta_{2}H^{2}\oplus\theta_{2}\overline{z}\overline{\theta_{1}}%
\overline{H^{2}}\cap\overline{z}\theta_{2}\overline{\mathcal{H}\left(
\theta_{2}\right) }\hspace{0.2in}\text{(by Lemma \ref{hthetalemma})}\\
& =\theta_{2}H^{2}\oplus\theta_{2}\overline{z}\left( \overline{\theta_{1}%
}\overline{H^{2}}\cap\overline{\mathcal{H}\left( \theta_{2}\right) }\right)
\\
& =\theta_{2}H^{2}\oplus\theta_{2}\overline{z}\overline{\left( \theta
_{1}H^{2}\cap\mathcal{H}\left( \theta_{2}\right) \right) }\text{.}%
\end{align*}
\end{proof}
From this, we get:
\begin{corollary}
\label{rangef}For any two inner functions $\theta_{1}$ and $\theta_{2}$,
\[
\operatorname*{closure}\left( \operatorname*{range}\left( H_{\overline
{\theta_{2}}}^{\ast}H_{\overline{\theta_{1}}}\right) \right) =\theta
_{2}\overline{z}\overline{\left( \mathcal{H}\left( \theta_{2}\right)
\ominus\left( \theta_{1}H^{2}\cap\mathcal{H}\left( \theta_{2}\right)
\right) \right) }.
\]
\end{corollary}
\begin{proof}%
\begin{align*}
\operatorname*{closure}\left( \operatorname*{range}\left( H_{\overline
{\theta_{2}}}^{\ast}H_{\overline{\theta_{1}}}\right) \right) & =\left(
\ker H_{\overline{\theta_{1}}}^{\ast}H_{\overline{\theta_{2}}}\right)
^{\perp}\\
& =\left( \theta_{2}H^{2}\oplus\theta_{2}\overline{z}\overline{\left(
\theta_{1}H^{2}\cap\mathcal{H}\left( \theta_{2}\right) \right) }\right)
^{\perp}\\
& =\mathcal{H}\left( \theta_{2}\right) \ominus\theta_{2}\overline
{z}\overline{\left( \theta_{1}H^{2}\cap\mathcal{H}\left( \theta_{2}\right)
\right) }\\
& =\theta_{2}\overline{z}\overline{\mathcal{H}\left( \theta_{2}\right)
}\ominus\theta_{2}\overline{z}\overline{\left( \theta_{1}H^{2}\cap
\mathcal{H}\left( \theta_{2}\right) \right) }\\
& =\theta_{2}\overline{z}\overline{\left( \mathcal{H}\left( \theta
_{2}\right) \ominus\left( \theta_{1}H^{2}\cap\mathcal{H}\left( \theta
_{2}\right) \right) \right) }\text{.}%
\end{align*}
The question of when $\ker H_{\overline{\theta_{1}}}^{\ast}H_{\overline
{\theta_{2}}}=\ker H_{\overline{\theta_{2}}}$ can now be answered as follows.
\end{proof}
\begin{corollary}
For inner functions $\theta_{1}$ and $\theta_{2}$, $\ker H_{\overline
{\theta_{1}}}^{\ast}H_{\overline{\theta_{2}}}=\ker H_{\overline{\theta_{2}}}$
if and only if $\theta_{1}H^{2}\cap\mathcal{H}\left( \theta_{2}\right)
=\left\{ 0\right\} $.
\end{corollary}
\begin{proof}
By Theorem \ref{hkp2}, $\ker H_{\overline{\theta_{1}}}^{\ast}H_{\overline
{\theta_{2}}}=\theta_{2}H^{2}\oplus\theta_{2}\overline{z}\overline{\left(
\theta_{1}H^{2}\cap\mathcal{H}\left( \theta_{2}\right) \right) }$, so $\ker
H_{\overline{\theta_{1}}}^{\ast}H_{\overline{\theta_{2}}}=\ker H_{\overline
{\theta_{2}}}=\theta_{2}H^{2}$ when $\theta_{1}H^{2}\cap\mathcal{H}\left(
\theta_{2}\right) =\left\{ 0\right\} $.
\end{proof}
We can use the results above to provide a different proof of the main theorem
of Nakazi from \cite{N},
\begin{theorem}
[Nakazi, 1987]Let $n$ be a nonnegative integer. Let $\theta_{1}$ and
$\theta_{2}$ be inner functions. Suppose that $\mathcal{H}\left( \theta
_{2}\right) \cap\theta_{1}H^{2}\neq\left\{ 0\right\} .$ Then the dimension
of $\mathcal{H}\left( \theta_{2}\right) /\left( \mathcal{H}\left(
\theta_{2}\right) \cap\theta_{1}H^{2}\right) $ is $n$ if and only if
$\theta_{1}$ is a Blaschke product of degree $n$.
\end{theorem}
\begin{proof}
The hypothesis that $\mathcal{H}\left( \theta_{2}\right) \cap\theta_{1}%
H^{2}\neq\left\{ 0\right\} $, by the corollary above, implies that $\ker
H_{\overline{\theta_{1}}}^{\ast}H_{\overline{\theta_{2}}}\neq\ker
H_{\overline{\theta_{2}}}$. By Theorem \ref{kernelthm}, $\ker H_{\overline
{\theta_{2}}}^{\ast}H_{\overline{\theta_{1}}}=\ker H_{\overline{\theta_{1}}},$
that is $\operatorname*{closure}\left( \operatorname*{range}\left(
H_{\overline{\theta_{1}}}^{\ast}H_{\overline{\theta_{2}}}\right) \right)
=\operatorname*{Range}\left( H_{\overline{\theta_{1}}}^{\ast}\right)
=\mathcal{H}\left( \theta_{1}\right) $.
By Corollary \ref{rangef}, $\dim\mathcal{H}\left( \theta_{2}\right) /\left(
\mathcal{H}\left( \theta_{2}\right) \cap\theta_{1}H^{2}\right) =n$ exactly
when
\[
\dim\left[ \operatorname*{closure}\left( \operatorname*{range}\left(
H_{\overline{\theta_{2}}}^{\ast}H_{\overline{\theta_{1}}}\right) \right)
\right] =\operatorname*{rank}\left( H_{\overline{\theta_{2}}}^{\ast
}H_{\overline{\theta_{1}}}\right) =n\text{,}%
\]
or, equivalently,
\[
\operatorname*{rank}\left( H_{\overline{\theta_{2}}}^{\ast}H_{\overline
{\theta_{1}}}\right) =\operatorname*{rank}\left( H_{\overline{\theta_{1}}%
}^{\ast}H_{\overline{\theta_{2}}}\right) =\dim\mathcal{H}\left( \theta
_{1}\right) =n.
\]
This happens exactly when $\theta_{1}$ is a Blaschke product of degree $n$.
\end{proof}
To understand better when $\ker H_{\overline{\theta_{1}}}^{\ast}%
H_{\overline{\theta_{2}}}=\ker H_{\overline{\theta_{2}}}$ and when $\ker
H_{\overline{\theta_{2}}}^{\ast}H_{\overline{\theta_{1}}}=\ker H_{\overline
{\theta_{1}}}$, we define a partial order on the set of all inner functions as follows:
\begin{definition}
$\theta_{1}\prec\theta_{2}$ if $\ker H_{\overline{\theta_{1}}}^{\ast
}H_{\overline{\theta_{2}}}\neq\ker H_{\overline{\theta_{2}}}$, i.e.,
$\theta_{1}H^{2}\cap\mathcal{H}\left( \theta_{2}\right) \neq\left\{
0\right\} $.
\end{definition}
This does, in fact, give us a partial order since if $\theta_{1}\prec
\theta_{2}$ and $\theta_{2}\prec\theta_{3}$, then there are functions $h_{1}$
and $h_{2}$ in $H^{2}$ such that $\theta_{1}h_{1}\in\mathcal{H}\left(
\theta_{2}\right) $ and $\theta_{2}h_{2}\in\mathcal{H}\left( \theta
_{3}\right) $. But then $\overline{\theta_{2}}\theta_{1}h_{1}\in
\overline{H_{0}^{2}}$, and $\overline{\theta_{3}}\theta_{2}h_{2}\in
\overline{H_{0}^{2}}$, so $\left( \overline{\theta_{3}}\theta_{2}%
h_{2}\right) \left( \overline{\theta_{2}}\theta_{1}h_{1}\right)
=\overline{\theta_{3}}\theta_{1}h_{1}h_{2}\in\overline{H_{0}^{2}}$, i.e.,
$\theta_{1}h_{1}h_{2}\in\mathcal{H}\left( \theta_{3}\right) $, or
$\theta_{1}\prec\theta_{3}$.
\begin{proposition}
\label{strict}The order $\prec$ is a strict partial order.
\end{proposition}
\begin{proof}
By Theorem \ref{kernelthm}, it is impossible to have both $\ker H_{\overline
{\theta_{1}}}^{\ast}H_{\overline{\theta_{2}}}\neq\ker H_{\overline{\theta_{2}%
}}$ and $\ker H_{\overline{\theta_{2}}}^{\ast}H_{\overline{\theta_{1}}}%
\neq\ker H_{\overline{\theta_{1}}}$, i.e., $\theta_{1}\prec\theta_{2}$ and
$\theta_{2}\prec\theta_{1}$.
\end{proof}
Exactly when $\theta_{1}\prec\theta_{2}$ is still not fully understood, but
there are several things we can say.
\begin{enumerate}
\item If $z\theta_{1}|\theta_{2}$, then $\theta_{1}\prec\theta_{2}$.
\begin{proof}
If $z\theta_{1}|\theta_{2}$, then $\overline{z\theta_{1}}\theta_{2}$ $\in
H^{2}$, so $\overline{\theta_{2}}\theta_{1}\in\overline{H_{0}^{2}}$, and thus
$\theta_{1}\prec\theta_{2}$.
\end{proof}
\item \label{fact}If $\deg\theta_{1}>\deg\theta_{2}$ for finite Blaschke
products $\theta_{1}$ and $\theta_{2}$, or for an infinite Blaschke product
$\theta_{1}$ (for which we say that $\deg\theta_{1}=\infty$) and a finite
Blaschke product $\theta_{2}$, then $\theta_{1}\nprec\theta_{2}$.
\begin{proof}
We have an explicit representation in these cases for $\mathcal{H}\left(
\theta_{2}\right) $ (see \cite[page 278]{CFAF}) which tells us that if
$\theta_{2}$ is a finite Blaschke product, then $\mathcal{H}\left( \theta
_{2}\right) $ contains members whose inner factors can be any finite Blaschke
product of degree smaller than $\deg\theta_{2}$, and if $\theta_{2}$ is an
infinite Blaschke product, then $\mathcal{H}\left( \theta_{2}\right) $
contains members whose inner factor is any desired finite Blaschke product.
\end{proof}
\item If $\deg\theta_{1}1,$ the number $1/\overline{\zeta}$ is a zero of
$z^{m}\varphi\left( z\right) $ in the open unit disk of multiplicity greater
than or equal to the multiplicity of $\zeta.$
\end{theorem}
We next show how to derive this generalization also from Theorem \ref{circ1}
by using our notation and Cowen's characterization. It follows from the
assumption that
\[
z^{m}\varphi\left( z\right) =p(z)q(z)
\]
where $p(z)$ is the polynomial of degree $N+1$ whose zeros consist of $N+1$
elements from $\Im$ such that if $\zeta$ is a zero then $1/\overline{\zeta}$
is also a zero of $p(z).$ Thus $q(z)$ is a polynomial of degree $m-1.$ It is
easy to verify that $p(z)$ satisfies the relation
\[
p^{\sharp}(z):=z^{N+1}\overline{p(z)}=p(z).
\]
We note that the previous example is a special case of this example since by
(\ref{circu}), $z^{m}\varphi\left( z\right) \ =(z^{N+1}+e^{i\omega})g(z)$.
By a variant of Cowen's Theorem, as noted in \cite{nt}, $T_{\varphi}$ is
hyponormal if and only if there is some $k\in H^{\infty}$, $\left\|
k\right\| _{\infty}\leq1$ and $h\in H^{2}$ with
\[
\varphi-k\overline{\varphi}=h.
\]
Multiplying both sides by $z^{m}$, we have
\[
p(z)q(z)-kz^{N+1}\overline{p(z)}z^{m-N-1}\overline{q(z)}=p(z)\left[
q(z)-kz^{m-N-1}\overline{q(z)}\right] =z^{m}h.
\]
Equivalently
\[
\overline{z}^{m}q(z)-k\overline{z^{N+1}q(z)}=h_{1}%
\]
for some $h_{1}\in H^{2}.$ By Cowen's Theorem again, this happens if and only
if $T_{\psi}$ is hyponormal where
\[
\psi=z^{N+1}q(z)+\overline{z}^{m}q(z)=f(z)+\overline{z}^{N+1+m}%
f(z)=f(z)+\overline{\theta}f(z).
\]
with $\theta=z^{N+1+m}.$ Note that $\psi$ is of circulant type exactly as
defined above. The result follows now from Theorem \ref{circ1} as explained above.
\section{Hyponormality of $T_{\varphi}$ with $\left\| \varphi_{+}\right\|
_{2}=\left\| \varphi_{-}\right\| _{2}\label{hypopsec}$}
Let $\varphi\in L^{2}$ of the unit circle, and write $\varphi=\varphi
_{+}+\overline{\varphi_{-}}$, where $\varphi_{+}$ is the analytic part of
$\varphi$, i.e., the projection of $\varphi$ onto $H^{2}$, and $\overline
{\varphi_{-}}$ is the anti-analytic part of $\varphi$, i.e., $\varphi_{-}\in
H_{0}^{2}$. In this section we characterize hyponormal Toeplitz operators
$T_{\varphi}$ with symbols $\varphi$ satisfying $\left\| \varphi_{+}\right\|
_{2}=\left\| \varphi_{-}\right\| _{2}.$ As a consequence we get a
strengthened version of Theorem \ref{circ1}.
\begin{theorem}
For $\varphi=\varphi_{+}+\overline{\varphi_{-}}$ as above, if $\left\|
\varphi_{+}\right\| _{2}=\left\| \varphi_{-}\right\| _{2}$, then the
Toeplitz operator $T_{\varphi}$ is hyponormal if and only if $\varphi
_{-}|\varphi_{+}$ (in $H^{2}$).
\end{theorem}
\begin{proof}
By Cowen's Theorem, $T_{\varphi}$ is hyponormal if and only if there is some
$k\in H^{\infty}$ with $\left\| k\right\| _{\infty}\leq1$, and $h\in H^{2}$
with
\begin{equation}
\overline{\varphi_{-}}-k\overline{\varphi_{+}}=h\text{.}\label{C1}%
\end{equation}
When this happens, we have
\begin{align*}
\left\| \varphi_{-}\right\| _{2} & =\left\| \overline{\varphi_{-}%
}\right\| _{2}=\left\| P_{\overline{H_{0}^{2}}}\overline{\varphi_{-}%
}\right\| _{2}=\left\| P_{\overline{H_{0}^{2}}}k\overline{\varphi+}\right\|
_{2}\\
& \leq\left\| k\overline{\varphi+}\right\| _{2}\leq\left\| k\right\|
_{\infty}\left\| \overline{\varphi+}\right\| _{2}\leq\left\| \varphi
_{+}\right\| _{2}=\left\| \varphi_{-}\right\| _{2}\text{.}%
\end{align*}
We can thus conclude that we have equality everywhere in the above, in
particular, that $k$ must be an inner function, and $\left\| P_{\overline
{H_{0}^{2}}}k\overline{\varphi+}\right\| _{2}=\left\| k\overline{\varphi
+}\right\| _{2}$, i.e., that $k\overline{\varphi+}\in\overline{H_{0}^{2}}$,
so $\overline{\varphi_{-}}-k\overline{\varphi_{+}}\in\overline{H_{0}^{2}}$,
and thus $h=0$. We then have
\[
\overline{\varphi_{-}}=k\overline{\varphi_{+}}\text{.}%
\]
Since $k$ is an inner function, we can write
\[
k\varphi_{-}=\varphi_{+}\text{,}%
\]
or $\varphi_{-}|\varphi_{+}$.
In the opposite direction, it is clear that if $\varphi_{-}|\varphi_{+}$, then
$\theta\varphi_{-}=\varphi_{+}$, and (\ref{C1}) holds, with $h=0$, and
$k=\theta$.
\end{proof}
\begin{corollary}
\label{factor}For $\varphi=\varphi_{+}+\overline{\varphi_{-}}$ as above, if
$\left\| \varphi_{+}\right\| _{2}=\left\| \varphi_{-}\right\| _{2}$, then
the Toeplitz operator $T_{\varphi}$ is hyponormal if and only if
$\varphi=k\overline{\varphi}$ for some inner function $k$.
\end{corollary}
\begin{proof}
$T_{\varphi}$ is hyponormal$\ $iff $k\varphi_{-}=\varphi_{+}$ for some inner
function $k$, by the proof above. For $\varphi=\varphi_{+}+\overline
{\varphi_{-}}=k\varphi_{-}+\overline{\varphi_{-}}$, $\overline{\varphi
}=\overline{k}\overline{\varphi_{-}}+\varphi_{-}$, so $k\overline{\varphi
}=\theta\left( \overline{k}\overline{\varphi_{-}}+\varphi_{-}\right)
=k\varphi_{-}+\overline{\varphi_{-}}=\varphi$.
\end{proof}
\begin{corollary}
If $\varphi=f+\overline{\theta}f$, with notation as in (\ref{notation}) and
(\ref{notation2}), then $T_{\varphi}$ is hyponormal if and only if $\theta
_{1}|\theta_{2}$.
\end{corollary}
\begin{proof}
$\varphi_{+}=f$, and $\varphi_{-}=f^{\sharp}$. $\left\| \varphi_{+}\right\|
_{2}=\left\| \varphi_{-}\right\| _{2}$, so $T_{\varphi}$ is hyponormal if
and only $\varphi_{-}|\varphi_{+}$, i.e., $f^{\sharp}|f$, or $\theta
_{1}|\theta_{2}$.
\end{proof}
This generalizes Theorem \ref{circ1}, since we no longer need the condition
that $\theta\nprec\theta_{1}\theta_{2}$ in order to get the conclusion that
$T_{\varphi}$ is hyponormal if and only if $\theta_{1}|\theta_{2}$.
If $\varphi\in L^{2}$ satisfies $\left\| \varphi_{+}\right\| _{2}=\left\|
\varphi_{-}\right\| _{2}$ and $T_{\varphi}$ is hyponormal, then, as noted in
Corollary \ref{factor}, we can write, $k\varphi_{-}=\varphi_{+}$ for some
inner function $k$. This means that $\varphi_{+}$ and $\varphi_{-}$ have the
same outer factor, say $f$, and we can write $\varphi_{-}=\theta_{1}f$ and
$\varphi_{+}=k\theta_{1}f$ for some inner function $\theta_{1}$. The general
form of the symbol $\varphi$ of a hyponormal Toeplitz operator $T_{\varphi}$
for which $\left\| \varphi_{+}\right\| _{2}=\left\| \varphi_{-}\right\|
_{2}$ is
\[
\varphi=\varphi_{+}+\overline{\varphi_{-}}=k\theta_{1}f+\overline{\theta_{1}%
f}.
\]
One consequence of the above is that if $\left\| \varphi_{+}\right\|
_{2}=\left\| \varphi_{-}\right\| _{2}$, and $T_{\varphi}$ is hyponormal,
then, since $\varphi_{-}|\varphi_{+}$, we must have $\varphi_{+}\in H_{0}^{2}%
$. If $\varphi=\varphi_{+}+c+\overline{\varphi_{-}}$ where $\varphi
_{+},\varphi_{-}\in H_{0}^{2}$, and $c$ is a constant, then it is easy to see
that $T_{\varphi}$ is hyponormal if and only if $T_{\varphi-c}=T_{\varphi
_{+}+\overline{\varphi_{-}}}$ is hyponormal, so if $\left\| \varphi
_{+}\right\| _{2}=\left\| \varphi_{-}\right\| _{2}$, then $T_{\varphi}$ is
hyponormal if and only if $\varphi_{-}|\varphi_{+}$.
\section{Rank of $\left[ T_{\varphi}^{\ast},T_{\varphi}\right]
\label{ranksec}$}
It is relevant to mention the following characterization of hyponormal
Toeplitz operators whose self-commutator $\left[ T_{\varphi}^{\ast
},T_{\varphi}\right] $ is of finite rank \cite{nt}.\vspace{0.1in}
\noindent\textbf{Theorem (Nakazi and Takahashi, 1993).} $T_{\varphi}%
$\emph{\ is hyponormal and the \ self-commutator }$\left[ T_{\varphi}^{\ast
},T_{\varphi}\right] $\emph{\ is of finite rank if and only if there exists a
finite Blaschke product }$k$\emph{\ such that }$\varphi-k\overline{\varphi}\in
H^{\infty}$\emph{\ and moreover one can chose a }$k$\emph{\ such that the
degree of }$k$\emph{\ is equal to the rank of }$\left[ T_{\varphi}^{\ast
},T_{\varphi}\right] .\vspace{0.1in}$
It is possible to have different $k$ such that $\varphi-k\overline{\varphi}\in
H^{\infty}$, see \cite[page 252]{lee1} for a simple example. To estimate the
rank of $\left[ T_{\varphi}^{\ast},T_{\varphi}\right] $ using the above
result, one has to carefully find or argue for the right or unique finite
Blaschke product, as was done in \cite{FL}, \cite{lee1} and \cite{lee2}.
Here we offer a slightly simpler approach. Let $k$ be any inner function such
that $\varphi-k\overline{\varphi}\in H^{\infty}$(or $\overline{\varphi_{-}%
}-k\overline{\varphi_{+}}\in H^{2}$). We can write
\begin{align*}
\left[ T_{\varphi}^{\ast},T_{\varphi}\right] & =H_{\overline{\varphi}%
}^{\ast}H_{\overline{\varphi}}-H_{\varphi}^{\ast}H_{\varphi}=H_{\overline
{\varphi}}^{\ast}H_{\overline{\varphi}}-H_{k\overline{\varphi}}^{\ast
}H_{k\overline{\varphi}}\\
& =H_{\overline{\varphi_{+}}}^{\ast}H_{\overline{k}}H_{\overline{k}}^{\ast
}H_{\overline{\varphi_{+}}}\text{\hspace{0.2in}by Lemma \ref{fromgu1}.}%
\end{align*}
We thus have
\begin{align*}
\operatorname*{rank}\left[ T_{\varphi}^{\ast},T_{\varphi}\right] &
=\operatorname*{rank}H_{\overline{\varphi_{+}}}^{\ast}H_{\overline{k}%
}H_{\overline{k}}^{\ast}H_{\overline{\varphi_{+}}}=\operatorname*{rank}%
H_{\overline{\varphi_{+}}}^{\ast}H_{\overline{k}}\\
& =\min\left\{ \operatorname*{rank}H_{\overline{\varphi_{+}}}%
,\operatorname*{rank}H_{\overline{k}}\right\}
\end{align*}
by the theorem of Axler, Chang, and Sarason \cite{ACS}.
\subsection{Example}
If $T_{\varphi}$ is hyponormal and $\left\| \varphi_{+}\right\|
_{2}=\left\| \varphi_{-}\right\| _{2}$, then we must have $\varphi
_{+}=k\varphi_{-}$. It is then clear that $\operatorname*{rank}H_{\overline
{\varphi_{+}}}\geq\operatorname*{rank}H_{\overline{k}}$, so
\[
\operatorname*{rank}\left[ T_{\varphi}^{\ast},T_{\varphi}\right]
=\min\left\{ \operatorname*{rank}H_{\overline{\varphi_{+}}}%
,\operatorname*{rank}H_{\overline{k}}\right\} =\operatorname*{rank}%
H_{\overline{k}}%
\]
\subsection{Example}
If $\varphi=f+\overline{\theta}f$ and $T_{\varphi}$ is hyponormal, we have
\begin{align*}
\operatorname*{rank}\left[ T_{\varphi}^{\ast},T_{\varphi}\right] &
=\operatorname*{rank}\left[ T_{f+\overline{\theta}f}^{\ast},T_{f+\overline
{\theta}f}\right] \\
& =\min\left\{ \operatorname*{rank}H_{\overline{f+\overline{\theta}f}}^{\ast
},\operatorname*{rank}H_{\overline{u}}\right\} \\
& =\min\left\{ \operatorname*{rank}H_{\overline{f}},\operatorname*{rank}%
H_{\overline{u}}\right\} =\operatorname*{rank}H_{\overline{u}},
\end{align*}
where $u$ is the inner function defined by $f+\overline{\theta}f=u\overline
{f+\overline{\theta}f}$, or, equivalently, $uf^{\sharp}=f$, i.e., $u=\frac
{f}{f^{\sharp}}$. The fourth equality above comes from the fact that $u$ must
be a factor of $f$, and thus $\operatorname*{rank}H_{\overline{u}}%
\leq\operatorname*{rank}H_{\overline{f}}$.
In the example considered earlier, where $\varphi$ is a circulant polynomial,
$\frac{f}{f^{\sharp}}=z^{N-m}\frac{g\left( z\right) }{z^{m-1}\overline
{g\left( z\right) }}=u$, so we see that $u$ must be a Blaschke product with
degree $N-m+Z_{\mathbb{D}}-Z_{C\backslash\mathbb{D}}$, where $Z_{\mathbb{D}}$
is the number of zeros of $g$ in $\mathbb{D}$ and $Z_{C\backslash\mathbb{D}}$
is the number of zeros of $g$ outside of $\mathbb{D}$ (which is the number of
zeros of $z^{m-1}\overline{g\left( z\right) }$ in $\mathbb{D}$, all of which
cancel with the zeros of $g$ when $T_{\varphi}$ is hyponormal).
\subsection{Example}
In the more general case as in Theorem \ref{hkl}, let $k_{1}=\frac{\varphi
(z)}{\overline{\varphi(z)}}$. Write $k_{1}=z^{N-m}\frac{z^{m}\varphi(z)}%
{z^{N}\overline{\varphi\left( z\right) }}.$ \ Note that the zeros of
$z^{N}\overline{\varphi\left( z\right) }$ in $\mathbb{D}$ are the zeros of
$z^{m}\varphi(z)$ outside of $\mathbb{D}$. By Theorem \ref{hkl}, if
$T_{\varphi}$ is hyponormal, then for every zero $\zeta$ of $z^{m}%
\varphi\left( z\right) $ such that $\left| \zeta\right| >1,$ the number
$1/\overline{\zeta}$ is a zero of $z^{m}\varphi\left( z\right) $ in the open
unit disk of multiplicity greater than or equal to the multiplicity of $\zeta
$. We see that all the zeros of $z^{N}\overline{\varphi\left( z\right) }$ in
$\mathbb{D}$ will be canceled by the zeros of $z^{m}\varphi(z).$ Thus $k_{1}$
must be a Blaschke product with degree $N-m+Z_{\mathbb{D}}-Z_{C\backslash
\mathbb{D}}$, where $Z_{\mathbb{D}}$ is the number of zeros of $z^{m}%
\varphi(z)$ in $\mathbb{D}$ and $Z_{C\backslash\mathbb{D}}$ is the number of
zeros of $z^{m}\varphi(z)$ outside of $\mathbb{D}$ \ By the assumption that
$\Im$ contains at least ($N+1$) elements \ and the degree $z^{m}\varphi\left(
z\right) $ is $N+m$, it is clear that $Z_{\mathbb{D}}-Z_{C\backslash
\mathbb{D}}\leq m.$Therefore
\begin{align*}
\operatorname*{rank}\left[ T_{\varphi}^{\ast},T_{\varphi}\right] &
=\min\left\{ \operatorname*{rank}H_{\overline{\varphi_{+}}}%
,\operatorname*{rank}H_{\overline{k}_{1}}\right\} \\
& =\min\left\{ N,N-m+Z_{\mathbb{D}}-Z_{C\backslash\mathbb{D}}\right\}
=N-m+Z_{\mathbb{D}}-Z_{C\backslash\mathbb{D}}.
\end{align*}
\subsection{Estimates of Rank}
For more general $\varphi\in L^{\infty}$, if $\varphi$ is of bounded type (i.
e. quotient of two bounded analytic functions), we can still estimate the rank
of the self-commutator of $T_{\varphi}$ when $T_{\varphi}$ is hyponormal.
\begin{align*}
\left[ T_{\varphi}^{\ast},T_{\varphi}\right] & =H_{\overline{\varphi}%
}^{\ast}H_{\overline{\varphi}}-H_{\varphi}^{\ast}H_{\varphi}\\
& =H_{\overline{\varphi_{+}}}^{\ast}H_{\overline{\varphi_{+}}}-H_{\overline
{\varphi_{-}}}^{\ast}H_{\overline{\varphi_{-}}}\geq0.
\end{align*}
Since the closure of $\operatorname*{range}\left( H_{\overline{\varphi_{+}}%
}^{\ast}H_{\overline{\varphi_{+}}}\right) $ is $\mathcal{H}\left( \theta
_{+}\right) $, where $\theta_{+}$ is an inner function with $\ker
H_{\overline{\varphi_{+}}}=\theta_{+}H^{2}$, we can give an upper bound for
the rank of $\left[ T_{\varphi}^{\ast},T_{\varphi}\right] $ by the dimension
of this range, which is $\deg\theta_{+}$, if $\theta_{+}$ is a finite Blaschke
product. Because of the positivity of $H_{\overline{\varphi_{+}}}^{\ast
}H_{\overline{\varphi_{+}}}-H_{\overline{\varphi_{-}}}^{\ast}H_{\overline
{\varphi_{-}}}$, we can conclude that $\theta_{-}|\theta_{+}$, where
$\theta_{-}$ is an inner function with $\ker H_{\overline{\varphi_{-}}}%
=\theta_{-}H^{2}$. Let $\theta=\theta_{+}/\theta_{-}$. It then follows that
\[
\operatorname*{range}\left[ T_{\varphi}^{\ast},T_{\varphi}\right]
\supset\mathcal{H}\left( \theta_{+}\right) \ominus\mathcal{H}\left(
\theta_{-}\right) =\mathcal{H}\left( \theta\theta_{-}\right) \ominus
\mathcal{H}\left( \theta_{-}\right) =\theta_{-}\mathcal{H}\left(
\theta\right)
\]
since $\mathcal{H}\left( \theta\theta_{-}\right) =\mathcal{H}\left(
\theta_{-}\right) \oplus\theta_{-}\mathcal{H}\left( \theta\right) $. Thus
we have $\operatorname*{rank}\left[ T_{\varphi}^{\ast},T_{\varphi}\right]
\geq\dim\left( \theta_{-}\mathcal{H}\left( \theta\right) \right)
=\deg\theta$, if $\theta=\theta_{+}/\theta_{-}$ is a finite Blaschke product.
As an example, consider the case where $\varphi$ is any trigonometric
polynomial, $\varphi\left( z\right) =\sum_{n=-m}^{N}c_{n}z^{n}$, with
$T_{\varphi}$ hyponormal (and thus $N\geq m$). We will have $\deg\theta_{+}%
=N$, and $\deg\theta_{-}=m$, so the results above tell us that
\[
N-m\leq\operatorname*{rank}\left[ T_{\varphi}^{\ast},T_{\varphi}\right] \leq
N\text{,}%
\]
which is discussed and proved in \cite{FL} (the upper bound is known from
\cite{IW}).
\section{More Hyponormal Toeplitz Operators\label{moresec}}
In this section we reduce the determination of the hyponormality of Toeplitz
operators with symbols of bounded type to the computation of the norm of
Hankel operators via the Nehari distance formula. This allows us to construct
a class of hyponormal Toeplitz operators with symbols satisfying certain
symmetric conditions, including those of polynomial ones discussed in
\cite{flt} and \cite{lee2}. This is achieved by using recent results on the
efficient computation of the norm of Hankel operators in the context of robust
control theory, which will be given in the next section.
Let $\varphi=\varphi_{+}+\overline{\varphi_{-}}$, as earlier, and assume
$\varphi$ is of bounded type. Write $\ker H_{\overline{\varphi_{-}}}=\theta
H^{2}$ for some inner function $\theta$, and, as noted above, the
hyponormality of $T_{\varphi}$ implies that there is some inner function
$\theta_{0}$ such that $\ker H_{\overline{\varphi_{+}}}=\theta_{0}\theta
H^{2}$. Then we can write
\[
\varphi_{+}=\theta_{0}\theta\overline{a},\hspace{0.2in}\varphi_{-}%
=\theta\overline{b}%
\]
for $a\in\mathcal{H}\left( \theta_{0}\theta\right) $ and $b\in
\mathcal{H}\left( \theta\right) $.
By Cowen's Theorem, $T_{\varphi}$ will be hyponormal if and only if there are
$k\in H^{\infty}$ with $\left\| k\right\| _{\infty}\leq1$ and $h\in H^{2}$
with
\[
\overline{\varphi_{-}}-k\overline{\varphi_{+}}=h
\]
or
\begin{align*}
\overline{\theta}b-k\overline{\theta_{0}\theta}a & =h\text{,}\\
\theta_{0}b-ka & =\theta_{0}\theta h\text{.}%
\end{align*}
If there is a solution to this last equation, then we must have $\theta_{0}%
|k$, so let $k=\theta_{0}k_{1}$ and write the equation as
\begin{equation}
b-k_{1}a=\theta h.\label{fneq}%
\end{equation}
Let $k_{1}\in H^{\infty}$ be any solution of equation (\ref{fneq}), i.e.,
$b-k_{1}a=\theta h$ for some $h\in H^{2}$. We do not place any restriction on
$\left\| k_{1}\right\| _{\infty}$. Then all solutions $g$ to $b-ga=\theta h$
are of the form $g=k_{1}+\theta f$ for some $f\in H^{\infty}$. We can thus
conclude that $T_{\varphi}$ is hyponormal iff there is some solution $g$ of
$H^{\infty}$ norm at most 1, or $\inf_{f\in H^{\infty}}\left\| k_{1}+\theta
f\right\| _{\infty}\leq1.$ By Nehari's Theorem \cite{nehari} or the result in
\cite{sarason}
\[
\inf_{f\in H^{\infty}}\left\| k_{1}+\theta f\right\| _{\infty}=\inf_{f\in
H^{\infty}}\left\| \overline{\theta}k_{1}+f\right\| _{\infty}=\left\|
H_{\overline{\theta}k_{1}}\right\| .
\]
Thus we have
\begin{theorem}
\label{hankel}If $\varphi=\theta_{0}\theta\overline{a}+\overline{\theta}b $,
for $a\in\mathcal{H}\left( \theta_{0}\theta\right) $ and $b\in
\mathcal{H}\left( \theta\right) $, \ then $T_{\varphi}$ is hyponormal iff
for any particular solution $k_{1}\in H^{\infty}$ of equation (\ref{fneq}),
$\left\| H_{\overline{\theta}k_{1}}\right\| \leq1$.
\end{theorem}
Let us now decompose $a$ as
\[
a=P_{\mathcal{H}\left( \theta\right) }a+\theta a_{1}=a_{0}+\theta
a_{1}\text{.}%
\]
The component $a_{1}$ does not have any effect on $\left\| H_{\overline
{\theta}k_{1}}\right\| $ for solutions $k_{1}$ of (\ref{fneq}), so we get
\begin{corollary}
If $\varphi=\theta_{0}\theta\overline{a}+\overline{\theta}b$, for
$a\in\mathcal{H}\left( \theta_{0}\theta\right) $ and $b\in\mathcal{H}\left(
\theta\right) $, with $a=P_{\mathcal{H}\left( \theta\right) }a+\theta
a_{1}=a_{0}+\theta a_{1}$, then the hyponormality of $T_{\varphi}$ is
independent of the component $a_{1}$.
\end{corollary}
\begin{corollary}
If $\varphi=\theta_{0}\theta\overline{a}+\overline{\theta}b$, for
$a\in\mathcal{H}\left( \theta_{0}\theta\right) $ and $b\in\mathcal{H}\left(
\theta\right) $, \ then $T_{\varphi}$ is hyponormal iff $T_{\psi}$ is
hyponormal, where $\psi=T_{\overline{\theta_{0}}}\theta_{0}\theta\overline
{a}+\overline{\theta}b=P(\theta\overline{a})+\overline{\theta}b=\theta
\overline{a_{0}}+\overline{\theta}b.$
\begin{proof}
Note that the hyponormality of both $T_{\varphi}$ and $T_{\psi}$ is equivalent
to $\left\| H_{\overline{\theta}k_{1}}\right\| \leq1$ where $k_{1}$ is any
bounded analytic solution of $b-k_{1}a_{0}=\theta h$ for some $h\in H^{2}.$
\end{proof}
\end{corollary}
Assume now that the inner function $\theta$ (as used in this section) has some
Blaschke factor, i.e., is zero at some $\alpha\in\mathbb{D}$, and that
$\left| a(\alpha)\right| =\left| b(\alpha)\right| $. If $T_{\varphi}$ is
hyponormal, i.e.,
\[
b-k_{1}a=\theta h
\]
has a solution with $\left\| k_{1}\right\| _{\infty}\leq1$, we then have
\[
b\left( \alpha\right) -k_{1}\left( \alpha\right) a\left( \alpha\right)
=\theta\left( \alpha\right) h\left( \alpha\right) =0,
\]
or $k_{1}\left( \alpha\right) =\frac{b\left( \alpha\right) }{a\left(
\alpha\right) }$ will have modulus one, which means that $k_{1}(z)$ must be
constant, by the maximum modulus principle, say $k_{1}(z)=\zeta$. We now
write
\[
b-\zeta\left( a_{0}+\theta a_{1}\right) =\theta h,
\]
and deduce that since $b-\zeta a_{0}\in\mathcal{H}\left( \theta\right) $,
$b-\zeta a_{0}=0$. This then gives us
\begin{corollary}
\label{equal}If $\varphi=\theta_{0}\theta\overline{a}+\overline{\theta}b$, for
$a\in\mathcal{H}\left( \theta_{0}\theta\right) $ and $b\in\mathcal{H}\left(
\theta\right) $, and $\left| a(\alpha)\right| =\left| b(\alpha)\right| $
for some $\alpha$ with $\theta\left( \alpha\right) =0$, then $T_{\varphi}$
is hyponormal iff $b=\zeta a_{0}$ for some constant $\zeta$ of modulus 1.
\end{corollary}
\subsection{Examples}
Consider again the case where $\varphi$ is any trigonometric polynomial,
$\varphi\left( z\right) =\sum_{n=-m}^{N}c_{n}z^{n}$, with $N\geq m$. We can
write $\varphi=\theta_{0}\theta\overline{a}+c_{0}+\overline{\theta}b$ where
$\theta(z)=z^{m}$, $\theta_{0}(z)=z^{N-m}$, $a(z)=c_{N}+c_{N-1}z+\cdots
+c_{1}z^{N-1}$, and $b(z)=c_{-1}z^{m-1}+c_{-2}z^{m-2}+\cdots+c_{-m}$. When we
write $a=P_{\mathcal{H}\left( \theta\right) }a+\theta a_{1}=a_{0}+\theta
a_{1}$ as above, we see that $a_{0}(z)=c_{N}+c_{N-1}z+\cdots+c_{N-m+1}z^{m-1}%
$, and $a_{1}(z)=c_{N-m}z^{m}+c_{N-m-1}z^{m+1}+\cdots+c_{1}z^{N-1}$. We thus
see that the hyponormality of $T_{\varphi}$ is independent of $c_{0}$, and,
since it is independent of $a_{1}$, it is independent of the Fourier
coefficients $c_{1},\ldots,c_{N-m}$ as well. This is a result demonstrated in
a different way in \cite[Theorem 1]{FL}.
Furthermore $T_{\varphi}$ is hyponormal if and only if $T_{\psi}$ is
hyponormal where $\psi(z)=T_{\overline{z^{N-m}}}z^{N}\overline{a}%
+\overline{z^{m}}b=z^{m}\overline{a_{0}\left( z\right) }+\overline{z}^{m}b,$
which is obtained in \cite[Proposition 2]{lee2}. This result shows that to
determine the hyponormality of $T_{\varphi}$ with $\varphi\left( z\right) $
of the form $\varphi\left( z\right) =\sum_{n=-m}^{N}c_{n}z^{n}$ $\left(
m\leq N\right) $, it is sufficient\ to consider the Toeplitz operator
$T_{\psi}$ with $\psi(z)$ of the form $\psi\left( z\right) =\sum_{n=-m}%
^{m}c_{n}z^{n}. $
Also, since $\theta(z)=z^{m}$, if $\left| a(0)\right| =\left| b(0)\right|
$, i.e., $\left| c_{N}\right| =\left| c_{-m}\right| $, and $T_{\varphi}$
is hyponormal, Corollary \ref{equal} says that we must have some constant
$\zeta$ of modulus one with $b=\zeta a_{0}$, i.e., $c_{-m+k}=\zeta c_{N-k}$
for $k=0,1,\ldots m-1$. This is a result which was shown in \cite[Theorem
1.4]{flt}.
\section{Norms of\ Hankel operators\label{normsec}}
In this section we give more explicit examples of hyponormal Toeplitz
operators with symbols satisfying some symmetric conditions including
polynomial symbols discussed in \cite{flt} and \cite{lee2}. We first introduce
some notation. Let $\varphi=\theta_{0}\theta\overline{a}+\overline{\theta}b$,
for $a\in\mathcal{H}\left( \theta_{0}\theta\right) $ and $b\in
\mathcal{H}\left( \theta\right) .$ Let also $\theta=\theta_{1}\theta_{2}$
where $\theta_{2}$ is a finite Blaschke product. Assume
\begin{equation}
b=\beta a+\theta_{1}b_{1}\text{ for some }b\in\mathcal{H}\left( \theta
_{2}\right) \text{ and constant }\beta\text{ (}\left| \beta\right|
\leq1).\label{assump}%
\end{equation}
Note that in this case equation (\ref{fneq}) becomes
\[
\beta a+\theta_{1}b_{1}-k_{1}a=\theta_{1}\theta_{2}h.
\]
This implies that $(\beta-k_{1})a=\theta_{1}h_{1}$ for some $h_{1}\in H^{2} $.
Therefore $k_{1}=\beta+\theta_{1}k_{2}$ where $k_{2}$ satisfies
\begin{equation}
b_{1}-k_{2}a=\theta_{2}h_{2}\label{ktwo}%
\end{equation}
for some $h_{2}\in H^{2}.$ \ By Theorem \ref{hankel}, $T_{\varphi}$,\ with
$\varphi$ satisfying (\ref{assump}), is hyponormal if and only if $\ \left\|
H_{\overline{\theta_{1}\theta_{2}}(\beta+\theta_{1}k_{2})}\right\| \leq1.$
The efficient computation of the norms of more general Hankel operators, of
the form $\left\| H_{\overline{\theta_{1}\theta_{2}}(\beta+\theta_{1}k_{2}%
)}\right\| $, where $\beta$ is a rational function (instead of a constant)
and $k_{2}$ is any function in $\ H^{\infty}$, has been studied extensively in
recent literature because of its important application in robust control
theory. We refer reader to the book \cite{foias}\ by Foias, \"{O}zbay and
Tannenbaum for details. In particular, if $\gamma=\left\| H_{\overline
{\theta_{1}\theta_{2}}(\beta+\theta_{1}k_{2})}\right\| >\gamma_{e}$, where
$\gamma_{e}=$ $\left\| H_{\overline{\theta_{1}\theta_{2}}(\beta+\theta
_{1}k_{2})}\right\| _{e}$ is the essential norm of $\ H_{\overline{\theta
_{1}\theta_{2}}(\beta+\theta_{1}k_{2})},$ then\ there exists an efficiently
computable matrix $K(\gamma)$ such that $\gamma$ is the largest value for
which $K(\gamma)$ is singular. The remarkable fact is that $K(\gamma)$ is of
order at most $2(n+l)$ where $n$ is the degree of $\beta$ and $l$ is the
degree of the finite Blaschke product $\theta_{2}$. There are several
efficient algorithms to compute $K(\gamma)$; see the paper \cite{ozbay} by the
first named author, Toker and \ \"{O}zbay for one algorithm which uses the
realization form of the rational functions $\beta$ and $\theta_{2}.$
Furthermore $\gamma_{e}$ can be computed as shown in Foias and\ Tannenbaum
\cite{ft}.
We next give examples by further assuming $\theta_{2}=z.$ Thus \ we can take
$k_{2}$ satisfying (\ref{ktwo}) to be a constant. In other words, we will
compute $\gamma=\left\| H_{\overline{\theta}\overline{z}(1+\alpha\theta
)}\right\| $ where $\alpha$ is any constant. Since the results in
\cite{ozbay} are derived for Hankel operators on the Hardy space of the right
half plane for the much more general case, here we give a self-contained
discussion for the computation of $\left\| H_{\overline{\theta}\overline
{z}(1+\alpha\theta)}\right\| .$
Let $\gamma_{e}=$ $\left\| H_{\overline{\theta}\overline{z}(1+\alpha\theta
)}\right\| _{e}$ be the essential norm of $\ H_{\overline{\theta}\overline
{z}(1+\alpha\theta)}.$ First note that since $H_{\overline{\theta}\overline
{z}(1+\alpha\theta)}=H_{\overline{\theta}\overline{z}}+H_{\alpha\overline{z}}%
$, $H_{\overline{\theta}\overline{z}(1+\alpha\theta)}$ is a rank one
perturbation of $H_{\overline{\theta}\overline{z}}$, we know that
$H_{\overline{\theta}\overline{z}(1+\alpha\theta)}$ and $H_{\overline{\theta
}\overline{z}}$ have the same essential norm. If $\theta$ is a finite Blaschke
product, then $H_{\overline{\theta}\overline{z}(1+\alpha\theta)}$ is of finite
rank, so $\gamma_{e}=0.$ If $\theta$ has as a factor an infinite Blaschke
product or a singular inner function, then $\gamma_{e}=1,$ since
$H_{\overline{\theta}\overline{z}}^{\ast}H_{\overline{\theta}\overline{z}}$ is
the projection onto the infinite dimensional space $\mathcal{H}\left(
z\theta\right) .$
Assume now $\gamma=\left\| H_{\overline{\theta}\overline{z}(1+\alpha\theta
)}\right\| >\gamma_{e}$. Then $\gamma$ is the largest value such that there
exists a pair of vectors $x$ and $y_{0}$ in $H^{2}$ (called a Schmidt pair)
satisfying
\begin{align*}
H_{\overline{\theta}\overline{z}(1+\alpha\theta)}x & =J(I-P)\left[
\overline{\theta}\overline{z}(1+\alpha\theta)x\right] =\gamma y_{0},\\
H_{\overline{\theta}\overline{z}(1+\alpha\theta)}^{\ast}y_{0} & =P\left[
\overline{\theta}\overline{z}(1+\alpha\theta)J^{\ast}y_{0}\right] =\gamma x.
\end{align*}
Write $J^{\ast}y_{0}=\overline{z}\overline{y}$ where $y\in H^{2}$. The above
equation holds if and only if there exist $\phi_{1}$ and $\phi_{2}$ in $H^{2}$
such that
\begin{align}
\overline{\theta}\overline{z}(1+\alpha\theta)x+\phi_{1} & =\gamma\overline
{z}\overline{y},\label{xequa}\\
\theta z(1+\overline{\alpha}\overline{\theta})\overline{z}\overline
{y}+\overline{z}\overline{\phi_{2}} & =\gamma x.\label{yeq}%
\end{align}
It follows from above equations that
\[
\gamma\overline{\theta}\overline{z}x=(1+\overline{\alpha}\overline{\theta
})\overline{z}\overline{y}+\overline{\theta}\overline{z}^{2}\overline{\phi
_{2}}\in\overline{zH^{2}},\quad\gamma\theta\overline{y}=(1+\alpha
\theta)x+\theta z\phi_{1}\in H^{2}.
\]
Therefore
\begin{equation}
\phi_{1}=-P\left[ \overline{\theta}\overline{z}(1+\alpha\theta)x\right]
=-P(\overline{\theta}\overline{z}x)-\alpha P(\overline{z}x)=-\alpha
\overline{z}\left[ x-x(0)\right] \label{newkey}%
\end{equation}
and
\begin{align}
\overline{z}\overline{\phi_{2}} & =-(I-P)\left[ \theta z(1+\overline{\alpha
}\overline{\theta})\overline{z}\overline{y}\right] \nonumber\\
& =-(I-P)\left[ \theta\overline{y}\right] -\overline{\alpha}(I-P)\left[
\overline{y}\right] =-\overline{\alpha}\left[ \overline{y}-\overline
{y(0)}\right] .\label{phit}%
\end{align}
Plugging (\ref{newkey})\ \ and \ (\ref{phit}) into \ (\ref{xequa})\ \ and
\ (\ref{yeq}), we get
\begin{align*}
\overline{\theta}x+\alpha x(0) & =\gamma\overline{y},\\
\theta\overline{y}+\overline{\alpha}\overline{y(0)} & =\gamma x.
\end{align*}
Solving $x$ and $\overline{y}$, we have
\begin{align*}
\left( \gamma^{2}-1\right) \overline{y} & =\overline{\alpha}\overline
{\theta}\overline{y(0)}+\gamma\alpha x(0),\\
\left( \gamma^{2}-1\right) x & =\alpha\theta x(0)+\gamma\overline{\alpha
}\overline{y(0)}.
\end{align*}
Therefore there exist\ non-zero solutions $x$ and $\overline{y}$ satisfying
the above equations if and only \ if there exist non-zero numbers\ $\ x(0)$
and $\overline{y}(0)$ satisfying
\begin{align*}
\left( 1+\overline{\alpha}\overline{\theta}(0)-\gamma^{2}\right)
\overline{y(0)}+\gamma\alpha x(0) & =0\\
\left( 1+\alpha\theta(0)-\gamma^{2}\right) x(0)+\gamma\overline{\alpha
}\overline{y(0)} & =0
\end{align*}
This happens if and only if the $2\times2$ matrix
\[
\ K(\gamma)=\left(
\begin{array}
[c]{cc}%
1+\overline{\alpha}\overline{\theta}(0)-\gamma^{2} & \gamma\alpha\\
\gamma\overline{\alpha} & 1+\alpha\theta(0)-\gamma^{2}%
\end{array}
\right)
\]
is singular. Equivalently
\[
\left| 1+\alpha\theta(0)-\gamma^{2}\right| ^{2}-\left| \gamma\alpha\right|
^{2}=0.
\]
Solving the above equation, we see that \ $\gamma=\left\| H_{\overline
{\theta}\overline{z}(1+\alpha\theta)}\right\| $ is given by
\[
\gamma^{2}=\frac{2+2\operatorname{Re}\left[ \alpha\theta(0)\right] +\left|
\alpha\right| ^{2}+\sqrt{4\left| \alpha\right| ^{2}(1+2\operatorname{Re}%
\left[ \alpha\theta(0)\right] )+\left| \alpha\right| ^{4}-4\left|
\operatorname{Im}\left[ \alpha\theta(0)\right] \right| ^{2}}}{2}%
\]
\begin{theorem}
\label{norm}Let $\varphi=z\theta\overline{a}+\overline{z}\left(
\overline{\theta}\beta a+a(0)\delta\right) $, for $a\in\mathcal{H}\left(
z\theta\right) $ and some constants $\delta$ and $\beta$ ($\left|
\beta\right| \leq1$ ). Then \ $T_{\varphi}$ is hyponormal if and only if
$\ \left\| H_{\overline{\theta}\overline{z}(\beta+\delta\theta)}\right\|
\leq1,$ or for $\alpha=\delta/\beta,$
\[
\frac{2+2\operatorname{Re}\left[ \alpha\theta(0)\right] +\left|
\alpha\right| ^{2}+\sqrt{4\left| \alpha\right| ^{2}(1+2\operatorname{Re}%
\left[ \alpha\theta(0)\right] )+\left| \alpha\right| ^{4}-4\left|
\operatorname{Im}\left[ \alpha\theta(0)\right] \right| ^{2}}}{2}\leq
\frac{1}{\left| \beta\right| ^{2}}.
\]
\end{theorem}
\begin{proof}
If we let $b=\beta a+a(0)\delta\theta,$ equation (\ref{ktwo}) becomes
$a(0)\delta-k_{2}a=zh_{2}$ for some $h_{2}\in H^{2}.$ Thus we can take
$k_{2}=\delta.$ Therefore $T_{\varphi}$ is hyponormal if and only if
$\ \left\| H_{\overline{\theta}\overline{z}(\beta+\delta\theta)}\right\|
\leq1$ or $\left\| H_{\overline{\theta}\overline{z}(1+\alpha\theta)}\right\|
\leq1/\left| \beta\right| $ for $\alpha=\delta/\beta.$ The result follows
now from above discussion.
\end{proof}
\subsection{Example}
Consider the case where $\varphi$ is any trigonometric polynomial,
$\varphi\left( z\right) =\sum_{n=-N}^{N}c_{n}z^{n}$. We can write
$\varphi=z\theta\overline{a}+c_{0}+\overline{z}\overline{\theta}b$ where
$\theta(z)=z^{N-1} $, $a(z)=c_{N}+c_{N-1}z+\cdots+c_{1}z^{N-1}$, and
$b(z)=c_{-1}z^{N-1}+c_{-2}z^{N-2}+\cdots+c_{-N}$. Assume $c_{-k}=\beta
c_{k\text{ }}$ for $k=2,\cdots N.$ That is $b(z)=\beta a(z)+c_{N}\delta
\theta(z)$, for $\delta=(c_{-1}-\beta c_{1})/c_{N}.$ $T_{\varphi}$ is
hyponormal if and only if
\[
2+\left| \frac{\delta}{\beta}\right| ^{2}+\sqrt{4\left| \frac{\delta}%
{\beta}\right| ^{2}+\left| \frac{\delta}{\beta}\right| ^{4}}\leq\frac
{2}{\left| \beta\right| ^{2}},
\]
or\ $\left| \delta\right| \leq1-\left| \beta\right| ^{2}$. Note that
$\beta=c_{-N}/c_{N}.$ So $T_{\varphi}$ is hyponormal if and only if
\[
\left| c_{-1}c_{N}-c_{1}c_{-N}\right| \leq\left| c_{N}\right| ^{2}-\left|
c_{-N}\right| ^{2}.
\]
The sufficiency of this condition is obtained in \cite[Theorem 1.8]{flt} while
the necessity of this condition is proved in \ \cite[Theorem 6]{lee2}.
We conclude the paper with some explicit examples of hyponormal Toeplitz
operators with irrational symbols. Let
\[
\varphi\left( z\right) =ze^{-\frac{1+z}{1-z}}+\beta\overline{z}%
e^{-\frac{1+\overline{z}}{1-\overline{z}}}+\beta\overline{z}%
\]
($\theta=e^{-\frac{1+z}{1-z}},a=1,b=\beta+\beta\theta,$ in the notation of
Theorem \ref{norm}).\ $T_{\varphi}$ is hyponormal if and only if
\[
\left| \beta\right| ^{2}\leq\frac{2}{3+2e^{-1}+\sqrt{5+8e^{-1}}}%
\approx.\,305\,15.
\]
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\end{document}